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Masteriza [31]
3 years ago
6

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

t=0 is v(0)=13. What is its position at time t=5?
Mathematics
1 answer:
user100 [1]3 years ago
3 0

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

a(t) = 24t + 2

Velocity:

v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

v(t) = 12t^{2} + 2t + 13

Position:

s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K

Since s(0) = 3

s(t) = 4t^{3} + t^{2} + 13t + 3

What is its position at time t=5?

This is s(5).

s(t) = 4t^{3} + t^{2} + 13t + 3

s(5) = 4*5^{3} + 5^{2} + 13*5 + 3

s(5) = 593

It's position at time t = 5 is 593.

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